how is the density and molar mass of gaseous substance related
Answers
(n/V) has the units of moles/liter. If we know the molecular mass of the gas, we can convert this into grams/liter(mass/volume). The molar mass (M) is the number of grams in one mole of a substance. If we multiply both sides of the above equation by the molar mass:
The left hand side is now the number of grams per unit volume, or the mass per unit volume (which is the density)Thus, the density (d) of a gas can be determined according to the following:
Alternatively, if the density of the gas is known, the molar mass of a gas can be determined:
Has the units of mass per unit volume
(n/V) has the units of moles/liter. If we know the molecular mass of the gas, we can convert this into grams/liter (mass/volume). The molar mass (M) is the number of grams in one mole of a substance. If we multiply both sides of the above equation by the molar mass:
The left hand side is now the number of grams per unit volume, or the mass per unit volume (which is the density)
Thus, the density (d) of a gas can be determined according to the following:
Alternatively, if the density of the gas is known, the molar mass of a gas can be determined:
Example:
What is the density of carbon tetrachloride vapor at 714 torr and 125°C?
The molar mass of CCl4 is 12.0 + (4*35.5) = 154 g/mol. 125°C in degrees Kelvin would be (273+125) = 398K. Since we are dealing with torr, the value of the gas constant, R, would be 62.36 L torr/mol K.
Caesar's Las Breath
Of the molecules in Caesar's last gasp, how many of them are in the breath you just took?
Given:
One breath= 2 liters of air at a pressure of 730 mmHg and 37 degrees Celsius.
Earth is a sphere with a radius of 6370 Km and an average barometric pressure of 760mm Hg. (D of Hg=13.6g/cm3)
The avg. molecular mass of air is 29 g/mol.
1. Number of moles in Caesar’s last breathe:
n = PV/RT = (0.96 atm)(2L)/(0.0821 L atm/mol K)(310 K)
n = 0.075 mol
2. Number of moles in the atmosphere:
a. Surface area of earth:
Area = (4)(p)(r2)
Area = 5.10 x 1014 square meters
b. Pressure of the atmosphere on the earth’s surface:
Pressure = 760 mm Hg = 1.01 x 105 Pascals = 1.01 x 105 Newtons/square meter
Pressure = 1.01 x 105 kg/m s2
c. Force of the atmosphere on the earth
Pressure = Force/Area
Therefore
Force = (Pressure)(Area)
Force = (1.01 x 105 kg/m s2)(5.10 x 1014 m2)
Force = 5.15 x 1019 kg m /s2
d. Mass of the atmosphere
Force = (mass)(acceleration)
therefore
mass = Force/acceleration
mass = (5.15 x 1019 kg m/s2)/(9.8 m/s2) note: this is the acceleration due to gravity
mass = 5.26 x 1018 kg or 5.26 x 1021 g
e. Moles in the atmosphere
mol = (5.26 x 1021 g)(1 mol/29 g)
mol = 1.81 x 1020 mol
3. Fraction of atmosphere which represents molecules from Caesar’s last breath:
(0.075 mol)/(1.81 x 1020 mol) = 4.14 x 10-22
4. Moles of Caesar’s last breath in your last breath:
Assume your breath holds 0.075 mol:
(0.075 mol)(4.14 x 10-22) = 3.11 x 10-23mol
5. Number of molecules:
(6.022 x 1023 molecules/mol)(3.11 x 10-23mol) = 18.7 molecules